It's transcendental, but it's a disguised version of an algebraic equation. When I referred to "trivial circumstances," I meant starting with an algebraic equation and "artificially" turning it into a transcendental one. Using $e^{\ln x} = x$ is what you did.
Doubt on underline 1 What do you mean by the edge subgraph of $G$ defined by $L$?IsIt a subgraph with all edges from $L$?
Doubt on underline 2 Why is the equality $|L \backslash M_H|=|L|-|M_H|$true? Why $|L\backslash M_H|\ge|U|$? Can you please explain?
@TedShifrin solving argument I get something like arctan(a/b)=+pi/2 . Now for b=0 I got circle equation . Also writing a>0 gives me second condition. Although I didn’t understand what wanted me to do
I have a question. Please confirm if my understanding is correct. If A and B are 2 events. And if A is a super-set of B (imagine a diagram B inside A). Then it means if B happens A also automatically happens. Correct so far ? So the probability of (A U B) is equal to probability of B
@AlexClark representation theory, probably undergrad but maybe grad if the students who told me the professor no longer as ridiculously harsh as he used to be are right, AT, AG, and a reading course in NT (not sure what exactly yet, will discuss with my professor soon)
Eh probably undergrad in any event, two grad classes + grad apps would be rough times
In fact, if you ever have a subgroup H of a group G, you automatically get a homomorphism from H into G (why?), which of course is only onto when H = G.
(I'm guessing you're working with groups and not rings or other algebraic structures for that matter.)
Suppose H is the given subgroup of the image K, and G is the codomain. When you say H 'is normal', do you mean H is normal as a subgroup of K or as a subgroup of G?
Suppose, as you figured I was thinking, that the statement is valid for the case of H being normal as a subgroup of K. Can you handle the other case, when H is normal as a subgroup of G?
@AlessandroCodenotti: Do you mind explaining a thing or two in your answer?
I don't get how you obtain $\bar{y}$ from $y_{2n}=x_{2n}$ and $y_{2n-1}=2^nx_{2n}$. If we start by plugging in $n=0$ in $y_{2n-1}$ to get $y_1$ we get $2^0x_{2\cdot0}=x_0$, but we have $y_1=x_1$. The rest of the sequence is off in a similar way. I'm also curious as of why $\overline{c_{00}}=\ell^1. I'm not that good with topology, so the closure of a set often seems a bit mysterious to me.
If $c\in E$ is defined as $c_{2n-1}=0$ for all $n\geq1$ and $c_{2n}=\frac{1}{2^n}$ for all $n\geq1$, and, again, $X=\{x=(x_n)_{n\geq1}\in E:x_{2n}=0\forall n\geq1\}$. Then isn't $c\cap X=\emptyset$? I was to show that $c\notin X+Y$, but then I am to construct $Z=X-c$ and check that $Z\cap Y=\emptyset$ where $Y=\{y=(y_n)_{n\geq1}\in E:y_{2n}=\frac{1}{2^n}y_{2n-1}\forall n\geq 1\}$. I'm a bit lost.
Okay. So it's wrong to think that if $c$ isn't in either $X$ or $Y$, then it isn't in $X+Y$. I guess that's just analogous as to sets of numbers. If $c$ isn't in $X=\{1,2,3\}$ and not in $Y=\{4,5,6\}$, then it still might be in $X+Y$.
@mercio Mhmm, but not every central simple algebra over $F$ must be a matrix algebra, must it? For example, if you tensor the real hamilton quaternions over some ring extension of $\mathbb{R}$, is it still central?
Ah, maybe you use the fact that every CSA is isomorphic to a matrix algebra over some field extension.
I think I found a solution. Seems like if $A$ is an algebra over a field $F$ and $F \subset R$ a ring extension, then $Z(A \otimes_F R) = Z(A) \otimes_F R$. From: mathoverflow.net/questions/137584/…
You mean that $X\cap Y=\emptyset$ and that if $z\in X+Y$ then $x\in X$ and $y\in Y$ are unique such that $z=x+y$ means $c$ can't be in $X+Y$ as there's only one choice for $x$ and $y$?
I mean to say that when you assume that $c$ is in $X+Y$ you don't know at first what $x$ and $y$ are possible in the decompositions of $c$ as $x+y$.
But it turns out that there is only one possibility, so you can deduce from there that $x$ has to be a certain sequence, and that $y$ has to be a certain sequence.
And form there you just have to realize that $x$ actually is not in $X$, or $y$ actually is not in $Y$
Well hmmm I don't want to get banned for not worshipping everything the internet says, all the time, according to the rules of what is correc but I find this interesting and want to discuss
well I can show you how ive got there tobias but from experience when people say something along the lines of that they arnt going to be helpful, and there is no point posting an actual question because its more of a series of conclusions the begin with a couple of PNT identities, one of them is the first one they use in the proof of Bertrand's postulate it's pretty long winded so im just waiting for me to get stuck on something significant really
Don't we just have that $x_7=c_7-y_7=\frac{1}{16}-16y_8$ and $x_8=c_8-y_8=-\frac{1}{16}y_7$ because of $y_{2n}=\frac{1}{2^n}y_{2n-1}\Leftrightarrow y_{2n-1}=2^ny_{2n}$?
nope only a very minor difference mercio that's why I find it very interesting, it can almost definitely be reduced to a much simpler statement but I need to get coke and food
the upper bound for the second double sum on the left hand side is $\pi(n+1)$ and on the right hand side of the arrow it's $n-1$
And... curse prime numbers, I still yet to comprehend these even though I now have a pretty solid idea on what $G_{\delta}$ sets, which is some of the most unvisualisable mathematical sets in existence, looks like in general
but I save that lengthly rambling later in another room
For now, there are only two important things I need to sort out: 1. Code the dynamic zoom in the number plotter so I can get to the 4th factorial decimal place of a liouville number 2. Organise the introduction of the thesis so at least I can call a day for the introduction
@AlessandroCodenotti: I just don't see how you get $\overline{y}=x_1,x_1,2x_2,x_2,4x_3,x_3,\dots$ from $y_{2n}=x_{2n}$ and $y_{2n-1}=2^n y_{2n}$. Just insert $n=1$. Okay. Then we have $y_1=2y_2$ and $y_2=x_2$, but we have $y_1=x_1$ and $y_2=x_1$. Why?
Hmm, the indices might be off by one, that's not really vital, the only important thing about $\overline{y}$ is that it agrees with $\overline{x}$ in all the even positions
I'm trying to prove that $(1+s)^n\geq 1+ns$ for $s>-1$, $n\in\mathbb{N}$. With recurrence, I proved it. But I'm trying to use the binomial expansion. The first two terms are indeed $1+ns$, so the goal is to show that the rest is positive or zero.
@AlessandroCodenotti Cool. I will be there for a conference at the end of November (well, at MPI, and I am not entirely sure what the relation is between that and the university).
So, every infinite subset, since we're working in $\mathbb{R}$ Coarsest is fewest. Finest is most. What about the opposite end? The finest for which every finite subspace has the trivial topology?
@Akiva the more open sets you have the more restrictive the statement "this net converges to a point" becomes, is finer or coarser a better description for this?
The restriction of a topology to a subset is trivial if every open set either contains the subset or does not intersect it. If you have an open set in R that is not R or the empty set, then there are points both inside and outside it. Take two such points and the restriction of the topology to those two is not trivial.
This means that the only topology trivial on finite sets is the trivial topology.
Doubt on underline 1 What do you mean by the edge subgraph of $G$ defined by $L$?IsIt a subgraph with all edges from $L$?
Doubt on underline 2 Why is the equality $|L \backslash M_H|=|L|-|M_H|$true? Why $|L\backslash M_H|\ge|U|$? Can you please explain?
